Is my working of this problem ok?

f:N->R

f(n) = $\displaystyle \frac{n^2+2log2n}{n+1}

$

prove f(n) = O(n) < - - - thats big O with a "-" in it

$\displaystyle |f(n) | = |\frac{n^2+2log2n}{n+1}|$

=$\displaystyle \frac{n^2+2+log2n}{n+1}$ since each term is >=0

<= $\displaystyle \frac{n^2+2n}{n+1}$ since log2n<=n

<=$\displaystyle \frac{n^2+2n^2}{n+1}$ since n>=1

= $\displaystyle \frac{3n^2}{n+1}$

=$\displaystyle \frac{3n}{1}$

=3n

$\displaystyle f(n) = \frac{n^2+2log2n}{n+1}$ since each term >= 0

>=$\displaystyle \frac{n^2}{n+1}$ since log2n is >= 0 for n>=1

= $\displaystyle \frac{n}{1}$

=n

Therfore

f(n) = O(n)

I really hope i am doing this correctly

Thanks for any help and input.